Least value of $\left|2017 z+\frac{1}{2018z}\right|$ for complex $z$ with $|z|\geq 2019$

Question

Let $z=x+iy$ be a complex number with $|z|\geq 2019$. What is the least value of $$\left|2017 z+\frac{1}{2018z}\right| \, ?$$

What I tried: Put $z=re^{i\alpha}$. Then $z^{-1}=\frac{1}{r}e^{-i\alpha}$ and $$ \left|2017 re^{i\alpha}+\frac{1}{2018r}e^{-i\alpha}\right| \\ = \left |\left(2017r+\frac{1}{2018r}\right)\cos \alpha+\left(2017r-\frac{1}{2018r}\right)\sin \alpha\right| \\ =\sqrt{\left|\left(2017r+\frac{1}{2018r}\right)^2\cos^2\alpha+\left(2017r-\frac{1}{2018r}\right)^2\sin^2\alpha\right|} $$

Answer 1

From the triangle inequality $$ |z+w| \ge |z| - |w| $$ (with equality if $\overline z w$ is zero or a negative real number) we get $$ \left|2017 z+\frac{1}{2018z}\right| \ge 2017 |z|-\frac{1}{2018|z|} $$ with equality if $z^2 \le 0$, that is for purely imaginary $z$. The right-hand side is increasing in $|z|$, so that the minimal value is obtained for the minimal $|z| = 2019$: $$ \left|2017 z+\frac{1}{2018z}\right| \ge 2017 \cdot 2019 -\frac{1}{2018 \cdot 2019} \, . $$ Equality holds for $z = \pm 2019 i$.

---

You can also continue with your approach (using polar coordinates) and substitute $\sin^2 \alpha = 1 - \cos^2 \alpha$:

$$ \sqrt{\left(2017r+\frac{1}{2018r}\right)^2\cos^2\alpha+\left(2017r-\frac{1}{2018r}\right)^2\sin^2\alpha} = \sqrt{ 4 \frac{2017}{2018}\cos^2\alpha+\left(2017r-\frac{1}{2018r}\right)^2} \ge 2017r-\frac{1}{2018r} $$ where the minimum is attained if $\cos^2 \alpha = 0$. The rest is as above.

Answer 2

Our objective function is given by $$f(x,y)= \left( {\frac {\sqrt {16583822760976}}{2018}\sqrt {{\frac { 1}{{x}^{2}+{y}^{2}} \left( {x}^{2}+ \left( y+{\frac{1}{2018}} \right) ^{2} \right) \left( {x}^{2}+ \left( y-{\frac{1}{2018}} \right) ^{2} \right) }}} \right) $$ with the condition $$\sqrt{x^2+y^2}\geq 2019$$ We get the Minimum for $$x=0,y=-2019$$